Chapter 2, Problem 24P

Maximizing Area A wire 10 cm long is cut into two pieces, one of length x and the other of length 10 − x, as shown in the figure. Each piece is bent into the shape of a square.

1. (a) Find a function that models the total area enclosed by the two squares.
2. (b) Find the value of x that minimizes the total area of the two squares.

To determine

To find: The function that models the total area enclosed by two squares.

Answer to Problem 24P

The function that models the area of both the squares is $A(x)=\frac{1}{8}\left(x^2-10x+50\right)$.

Explanation of Solution

Given:

Length of wire is $10\text{cm}$.

Length of first wire after cutting is x and length of second wire is $\left(10-x\right)$.

Calculation:

If length of wires are $x$ and $\left(10-x\right)$ then length of side of square is $\frac{x}{4}$ and $\left(\frac{10-x}{4}\right)$.

Total enclosed area $\left(A\right)$ by both the squares is,

$\left(A\right)=a_1+a_2$ (1)

Where, $A$ is total enclosed area, $a_1$ is area of square whose side is $\frac{x}{4}$ and $a_2$ is area of square whose side is $\left(\frac{10-x}{4}\right)$.

Area of a square is,

$\begin{array}{c}\text{Area}=\left(\text{side}\right)\cdot \left(\text{side}\right)\\ =l\cdot l\end{array}$ (2)

Where, l is side of square.

Substitute $\frac{x}{4}$ for l in above equation (2) for area $a_1$,

$\begin{array}{c}{a}_1=\frac{x}{4}\cdot \frac{x}{4}\\ =\frac{x^2}{16}\end{array}$

The area enclosed by square of side $\frac{x}{4}$ is $\frac{x^2}{16}$.

Substitute $\frac{10-x}{4}$ for $l$ in equation (2) for area $a_2$,

$\begin{array}{c}{a}_2=\left(\frac{10-x}{4}\right)\cdot \left(\frac{10-x}{4}\right)\\ =\frac{100-10x-10x+x^2}{16}\\ =\frac{x^2-20x+100}{16}\end{array}$

The total area is calculated as follows,

$\begin{array}{c}A\left(x\right)=\frac{x^2}{16}+\frac{x^2-20x+100}{16}\\ =\frac{2x^2-20x+100}{16}\\ =\frac{2\left(x^2-10x+50\right)}{16}\\ =\frac{x^2-10x+50}{8}\end{array}$

Further solving,

$A\left(x\right)=\frac{1}{8}\left(x^2-10x+50\right)$

Thus, the function that models the area of both the squares is $A(x)=\frac{1}{8}\left(x^2-10x+50\right)$.

To determine

To find: The value of $x$ such that area enclosed by the two squares is minimum.

Answer to Problem 24P

The value of x that minimizes the area of the square is 5 cm.

Explanation of Solution

Function that models the area of square as calculated in part (a) is $A(x)=\frac{1}{8}\left(x^2-10x+50\right)$.

Sketch the function $A\left(x\right)$ using graphing calculator as shown below.

From the above figure, it can be observed that the function has minimum value at $x=5$.

Therefore, for the area to be minimum the length of the first wire must be 5 cm.